Let be $C$ a non empty, compact and convex set and $S:=\mathbb{R}^n\setminus C$, where $n>1$. In many occasions our professor simply skips the part where we construct a path that connects two arbitrary point of $S$. I was wondering if it is possible to construct a curve that connects two such points of $S$ by the following idea:
As $C$ is bounded there exists a number $k\in\mathbb{R}$ sucht that $\Vert x\Vert_\infty \leq\frac{k}{2}$ for all $x\in C$. Let be $a,e\notin C$ and $a\neq e $.
1.) There must be a straight line that lies in $S$ which either connects $a$ with $(k,a_2,a_3,\dots,a_n)$ or $a$ with $(-k,a_2,a_3,\dots,a_n)$. Otherwise we can produce a contradiction regarding convexity of $C$. WLOG we assume it is $b:=(k,a_2,a_3,\dots,a_n)$ and define $L_1:[0,1]\to \mathbb{R}^n$ by $L_1(t):=a+t(b-a)$.
2.) There must be a straight line that lies in $S$ which either connects $e$ with $(e_1,k,e_3,\dots,e_n)$ or $e$ with $(e_1,-k,e_3,\dots,e_n)$. Otherwise we can produce a contradiction regarding convexity. WLOG we assume it is $d:=(e_1,k,e_3,\dots,e_n)$ and define $L_4:[0,1]\to \mathbb{R}^n$ by $L_4(t):=d+t(e-d)$.
3.) Let be $c:=(k,k,e_3,\dots,e_n)$ and consider $L_3:[0,1]\to\mathbb{R}^n$ given by $$ L_3(t):=c+t(d-c). $$ $L_3$ lies in $M$ as $\Vert x\Vert_\infty\geq k$ for all $x\in L_3([0,1])$.
4.) Let be $b:=(k,a_2,\dots,a_n)$ and $L_2:[0,1]\to\mathbb{R}^n$ with $$ L_2(t):=b+t(c-b), $$ which also lies in $M$.
The four line segments all lie in $M$ are and can be connected in such a way to form a path along $a\to b\to c\to d\to e$; in other words $L:[0,1]\to\mathbb{R}^n$ with $$ S(t):=\begin{cases} L_1(4t),&t\in\left[0,\frac{1}{4}\right]\\ L_2\left(4t-1\right),&t\in\left[\frac{1}{4},\frac{1}{2}\right]\\ L_3(4t-2),&t\in\left[\frac{1}{2},\frac{3}{4}\right]\\ L_4(4t-3),&t\in\left[\frac{3}{4},1\right] \end{cases} $$ connects $a$ with $e$ and lies in $S$.
Is this correct?